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Prove That: Cos 3 X Sin 3 X + Sin 3 X Cos 3 X = 3 4 Sin 4 X

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Question

Prove that:  \[\cos^3 x \sin 3x + \sin^3 x \cos 3x = \frac{3}{4} \sin 4x\]

 
Numerical
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Solution

\[\text{ We know } , \]
\[\cos3x = 4 \cos^3 x - 3\text{ cos } x\]
\[ \Rightarrow \cos^3 x = \frac{\cos3x + 3\text{ cos } x}{4} . . . \left( i \right)\]
\[\text{ Also } , \]
\[\sin3x = 3\text{ sin } x - 4 \sin^3 x\]
\[ \Rightarrow \sin^3 x = \frac{3\text{ sin } x - \sin3x}{4} . . . \left( ii \right)\]
\[Now, \]
\[LHS = \cos^3 x \sin3x + \sin^3 x \cos3x\]
\[ = \left( \frac{\cos3x + 3\text{ cos } x}{4} \right)\sin3x + \left( \frac{3\text{ sin } x - \sin3x}{4} \right)\cos3x\]
\[ \left[ \text{ Using  } \left( i \right) \text{ and }  \left( ii \right) \right]\]
\[ = \frac{1}{4}\left[ 3\left( \sin3x \text{ cos } x + \text{ sin } x \cos3x \right) + \left( \cos3x \sin3x - \sin3x \cos3x \right) \right]\]
\[ = \frac{1}{4}\left[ 3\sin\left( 3x + x \right) + 0 \right]\]
\[ = \frac{3}{4}\sin4x\]
\[ = RHS\]
\[\text{ Hence proved } .\]

 

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.2 [Page 36]

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R.D. Sharma Mathematics [English] Class 11
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.2 | Q 3 | Page 36

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